\documentclass[t,12pt,aspectratio=169]{beamer} % 16:9 宽屏比例，适合现代投影
\usepackage{ctex} % 中文支持
\usepackage{amsmath, amsthm, amssymb, bm} % 数学公式与符号
\usepackage{graphicx,tasks}
\usepackage{hyperref}
\usepackage{booktabs} % 用于高质量表格
\usepackage{tikz} % 可选：用于绘图
\usetikzlibrary{shapes,arrows,arrows.meta,3d, angles, quotes}
\usetikzlibrary{circuits.ee.IEC}
\usepackage{tikz-3dplot}

\addtobeamertemplate{frametitle}{}{\vspace*{0.7em}}

\usepackage{color}

% 设置段落间距
\usepackage{setspace}
%\onehalfspacing
\setlength{\parskip}{1em}  % 增加段落之间的间距为1em

% 主题设置（推荐简洁风格）
\usetheme{Madrid}
\usecolortheme{default} % 可选：seahorse, beaver, dolphin 等

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\title{高中数学第十一章 - 空间向量与立体几何}
\subtitle{随机练习}
\author{无名氏}

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\begin{document}

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% 标题页
\begin{frame}
  \titlepage
\end{frame}

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\begin{frame}[allowframebreaks]{11.E. }

问题：在平行六面体 $ABCD-A_1B_1C_1D_1$ 中，若直线 $AC$ 与 $BD$ 的交点为 $M$, 
设 $\overrightarrow{A_1B_1}=\boldsymbol{a}$, $\overrightarrow{A_1D_1}=\boldsymbol{b}$, $\overrightarrow{A_1A}=\boldsymbol{c}$, 
则下列向量中与 $B_1M$ 共线的向量是 (\quad). 

\begin{tasks}(1)
\task  $-2\boldsymbol{a}+2\boldsymbol{b}-\boldsymbol{c}$
\task  $\boldsymbol{a}+\boldsymbol{b}-2\boldsymbol{c}$
\task  $2\boldsymbol{a}-2\boldsymbol{b}-\boldsymbol{c}$
\task  $\boldsymbol{a}-\boldsymbol{b}-2\boldsymbol{c}$
\end{tasks}


% 设置观察角度，第一个参数是绕x轴旋转的角度，第二个是绕z轴旋转的角度
\tdplotsetmaincoords{40}{0} 

\begin{figure}[h]
    \centering
    \begin{tikzpicture}[scale=0.85, >=stealth, tdplot_main_coords]

        \coordinate (A) at (0,0,0);
        \coordinate (B) at (4,0,0);
        \coordinate (D) at ( 3.464, 2, 0); % (2*sqrt(3),2,0)
        \coordinate (A1) at (2.5, 1.443, 4.082); % (5/2, 5*sqrt(3)/6, 5*sqrt(6)/3)
        \coordinate (C) at (7.464,2,0);
        \coordinate (B1) at (6.5, 1.443, 4.082);
        \coordinate (C1) at (9.964, 3.443, 4.082);
        \coordinate (D1) at (5.964, 3.443, 4.082);

        \coordinate (M) at (3.732,1,0); %(A+C)/2
        % \coordinate (N) at (8.232, 2.443, 4.082); %(C1+B1)/2

        \draw[thick] (A) -- (B) -- (C);
        \draw[dashed] (C) -- (D) -- (A);
        \fill[blue!20, opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;

        \draw[thick] (A1) -- (B1) -- (C1) -- (D1) -- cycle;
        %\fill[blue!20, opacity=0.5] (A1) -- (B1) -- (C1) -- (D1) -- cycle;

        % 标记点
        \fill[blue] (A) circle (1.5pt) node[above left] {$A$};
        \fill[blue] (B) circle (1.5pt) node[below right] {$B$};
        \fill[blue] (C) circle (1.5pt) node[below right] {$C$};
        \fill[blue] (D) circle (1.5pt) node[above left] {$D$};
        \fill[blue] (A1) circle (1.5pt) node[below right] {$A_1$};
        \fill[blue] (B1) circle (1.5pt) node[below right] {$B_1$};
        \fill[blue] (C1) circle (1.5pt) node[below right] {$C_1$};
        \fill[blue] (D1) circle (1.5pt) node[below right] {$D_1$};
        \fill[blue] (M) circle (1.5pt) node[below left] {$M$};
        % \fill[blue] (N) circle (1.5pt) node[below right] {$N$};

        \draw[thick] (A) -- (A1);
        \draw[thick] (B) -- (B1);
        \draw[thick] (C) -- (C1);
        \draw[dashed] (D) -- (D1);

        \draw[dashed] (A) -- (C);
        \draw[dashed] (B) -- (D);

        % 基底
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (A1) -- (B1) node[midway, below] {$\mathbf{a}$};
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (A1) -- (D1) node[midway, above] {$\mathbf{b}$};
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (A1) -- (A) node[midway, left] {$\mathbf{c}$};

        % 待研究的线段
        \draw[->, -{Stealth[scale=1.5]}, dashed, blue] (B1) -- (M);

    \end{tikzpicture}
%    \caption{1.2-3}
%    \label{fig:1.2-3}
\end{figure}



\newpage

问题：已知$\vec{e}_1, \vec{e}_2$为单位向量，且$\vec{e}_1 \perp \vec{e}_2$. 若$\vec{a} = 2\vec{e}_1 + 3\vec{e}_2$，$\vec{b} = k\vec{e}_1 - 4\vec{e}_2$，$\vec{a} \perp \vec{b}$，则实数$k$的值为（\quad ）。

\begin{tasks}(1)
\task  3 \\
\task  $-3$ \\
\task  6 \\
\task  $-6$
\end{tasks}

解答：

已知 $\vec{e}_1, \vec{e}_2$ 为单位向量且 $\vec{e}_1 \perp \vec{e}_2$。  
给定 $\vec{a} = 2\vec{e}_1 + 3\vec{e}_2$，$\vec{b} = k\vec{e}_1 - 4\vec{e}_2$，且 $\vec{a} \perp \vec{b}$。  

由垂直条件，$\vec{a} \cdot \vec{b} = 0$：  
\[
(2\vec{e}_1 + 3\vec{e}_2) \cdot (k\vec{e}_1 - 4\vec{e}_2) = 0
\]  
展开得：  
\[
2k(\vec{e}_1 \cdot \vec{e}_1) - 8(\vec{e}_1 \cdot \vec{e}_2) + 3k(\vec{e}_2 \cdot \vec{e}_1) - 12(\vec{e}_2 \cdot \vec{e}_2) = 0
\]  
利用 $\vec{e}_1 \cdot \vec{e}_1 = 1$，$\vec{e}_2 \cdot \vec{e}_2 = 1$，$\vec{e}_1 \cdot \vec{e}_2 = 0$：  
\[
2k \cdot 1 - 12 \cdot 1 = 0 \implies 2k = 12 \implies k = 6.
\]  


\newpage

问题：已知空间四边形 $OABC$ 中，$OB = OC$，$\angle AOB = \angle AOC = \frac{\pi}{3}$，则 $\cos \langle \overrightarrow{OA}, \overrightarrow{BC} \rangle$ 的值为（\quad）。

\begin{tasks}(1)
\task $\frac{1}{2}$ 
\task $\frac{\sqrt{2}}{2}$ 
\task $-\frac{1}{2}$
\task $0$
\end{tasks}


\newpage

问题：在空间直角坐标系中 $Oxyz$ 中，已知 $A(2,-1,4)$, $B(-2,-1,-4)$. 则点A和点B关于 (\quad)。

\begin{tasks}(1)
\task $x$ 轴对称
\task 平面 $Oyz$ 对称
\task $y$ 轴对称
\task 平面 $Ozx$ 对称
\end{tasks}


% 设置观察角度，第一个参数是绕x轴旋转的角度，第二个是绕z轴旋转的角度
\tdplotsetmaincoords{60}{125} 

\begin{figure}[h]
    \centering
    \begin{tikzpicture}[scale=0.6, >=stealth, tdplot_main_coords]

        \coordinate (O) at (0,0,0);

        \coordinate (I) at (1,0,0);
        \coordinate (J) at (0,1,0);
        \coordinate (K) at (0,0,1);

        \coordinate (M) at (-4,-1,-4);
        \coordinate (N) at (-4,-1,4);
        \coordinate (P) at (4,-1,-4);
        \coordinate (Q) at (4,-1,4);
        \draw[dashed] (M) -- (N) -- (Q) -- (P) -- cycle;
        \fill[blue!20, opacity=0] (M) -- (N) -- (Q) -- (P) -- cycle;

        \coordinate (S) at (0,-6,0);    
        \coordinate (R) at (0,6,0);    
        \coordinate (T) at (0,-1,0);    

        \draw[dashed] (S) -- (R);

        \coordinate (A) at (2,-1,4);    
        \coordinate (B) at (-2,-1,-4);    
        \draw[dashed] (A) -- (B);

        % 标记点
        \fill[blue] (A) circle (2.5pt) node[above left] {$A(2,-1,4)$};
        \fill[blue] (B) circle (2.5pt) node[below right] {$B(-2,-1,-4)$};
        \fill[blue] (T) circle (2.5pt) node[above left] {};

        % 基底
        \draw[->, -{Stealth[scale=1]}, thick, purple] (O) -- (I) node[left] {$\mathbf{i}$};
        \draw[->, -{Stealth[scale=1]}, thick, purple] (O) -- (J) node[right] {$\mathbf{j}$};
        \draw[->, -{Stealth[scale=1]}, thick, purple] (O) -- (K) node[above] {$\mathbf{k}$};


    \end{tikzpicture}
%    \caption{1.2-3}
%    \label{fig:1.2-3}
\end{figure}



\newpage

% 设置观察角度，第一个参数是绕x轴旋转的角度，第二个是绕z轴旋转的角度
\tdplotsetmaincoords{70}{120} 

\begin{figure}[h]
    \centering
    \begin{tikzpicture}[scale=2.5, >=stealth, tdplot_main_coords]

        \coordinate (D) at (0,0,0);
        \coordinate (A) at (1,0,0);
        \coordinate (B) at (1,1,0);
        \coordinate (C) at (0,1,0);
        \coordinate (A1) at (1,0,1);
        \coordinate (B1) at (1,1,1);
        \coordinate (C1) at (0,1,1);
        \coordinate (D1) at (0,0,1);

        \draw[thick] (A) -- (B) -- (C);
        \draw[dashed] (C) -- (D) -- (A);
        \fill[blue!20, opacity=0.5] (A) -- (B) -- (C) -- (O) -- cycle;

        \draw[thick] (A1) -- (B1) -- (C1) -- (D1) -- cycle;
        %\fill[blue!20, opacity=0.5] (A1) -- (B1) -- (C1) -- (D1) -- cycle;

        % 标记点
        \fill[blue] (A) circle (1.0pt) node[below right] {$A$};
        \fill[blue] (B) circle (1.0pt) node[below right] {$B$};
        \fill[blue] (C) circle (1.0pt) node[below right] {$C$};
        \fill[blue] (D) circle (1.0pt) node[below right] {$D$};
        \fill[blue] (A1) circle (1.0pt) node[above left] {$A_1$};
        \fill[blue] (B1) circle (1.0pt) node[below right] {$B_1$};
        \fill[blue] (C1) circle (1.0pt) node[above right] {$C_1$};
        \fill[blue] (D1) circle (1.0pt) node[above right] {$D_1$};

        \draw[thick] (A) -- (A1);
        \draw[thick] (B) -- (B1);
        \draw[thick] (C) -- (C1);
        \draw[dashed] (D) -- (D1);

        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (A) -- (D1);
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (A) -- (B1);
%        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (C) -- (C1);
        \draw[->, -{Stealth[scale=1.5]}, dashed, purple] (D) -- (C1);
  
        \draw[->, -{Stealth[scale=1.5]}, thick, blue] (B1) -- (D1);

        %        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (B) --++ (1,0,-1);
 
    \end{tikzpicture}
%    \caption{1.3-8}
%    \label{fig:1.3-8}
\end{figure}







\end{frame}
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\end{document}
